15 moles of `H_(2)` and `5.2` moles of `I_(2)` are mixed are allowed to attain equilibrium at `500^(@)C`. At equilibrium the concentration of HI is found to be 10 moles. The equilibrium constant for the formation of HI is

Options

A. 50
B. 15
C. 100
D. 25

Explanation

To find the equilibrium constant \( K_c \) for the reaction between hydrogen and iodine to form hydrogen iodide, we can follow these steps:

### Step 1: Write the Balanced Chemical Equation
The reaction can be represented as:
\[ H_2(g) + I_2(g) \rightleftharpoons 2 HI(g) \]

### Step 2: Identify Initial Moles
From the problem, we have:
- Initial moles of \( H_2 = 15 \) moles
- Initial moles of \( I_2 = 5.2 \) moles
- At equilibrium, the moles of \( HI = 10 \) moles

### Step 3: Determine Change in Moles
Let \( x \) be the amount of \( H_2 \) that reacts. Since the stoichiometry of the reaction shows that 1 mole of \( H_2 \) reacts with 1 mole of \( I_2 \) to produce 2 moles of \( HI \), we can express the changes in moles as follows:
- Moles of \( H_2 \) at equilibrium: \( 15 - x \)
- Moles of \( I_2 \) at equilibrium: \( 5.2 - x \)
- Moles of \( HI \) at equilibrium: \( 2x \)

Given that at equilibrium \( 2x = 10 \), we can solve for \( x \):
\[ x = \frac{10}{2} = 5 \]

### Step 4: Calculate Moles at Equilibrium
Now we can substitute \( x \) back to find the equilibrium moles:
- Moles of \( H_2 \) at equilibrium:
\[ 15 - 5 = 10 \text{ moles} \]
- Moles of \( I_2 \) at equilibrium:
\[ 5.2 - 5 = 0.2 \text{ moles} \]

### Step 5: Write the Expression for \( K_c \)
The equilibrium constant \( K_c \) is given by the expression:
\[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]
Where \( [C] \) denotes the concentration of species \( C \).

### Step 6: Calculate Concentrations
Assuming the volume of the reaction vessel is \( V \):
- Concentration of \( HI \):
\[ [HI] = \frac{10}{V} \]
- Concentration of \( H_2 \):
\[ [H_2] = \frac{10}{V} \]
- Concentration of \( I_2 \):
\[ [I_2] = \frac{0.2}{V} \]

### Step 7: Substitute Values into \( K_c \) Expression
Now substituting the concentrations into the \( K_c \) expression:
\[ K_c = \frac{\left(\frac{10}{V}\right)^2}{\left(\frac{10}{V}\right)\left(\frac{0.2}{V}\right)} \]

### Step 8: Simplify the Expression
The \( V \) cancels out:
\[ K_c = \frac{100}{10 \times 0.2} = \frac{100}{2} = 50 \]

### Final Answer
Thus, the equilibrium constant \( K_c \) for the formation of \( HI \) is:
\[ \boxed{50} \]